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Question Number 195180 by mustafazaheen last updated on 26/Jul/23

$$$$$$1.\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{sinx},\frac{\pi }{2}<\mathrm{x}⩽2\pi \\ \mathrm{cosx},0⩽\mathrm{x}⩽\frac{\pi }{2}\end{array}$$$$\mathrm{then}\mathrm{find}\mathrm{the}{\mathrm{f}}^{{}^{\prime }}\left(\frac{\pi }{2}\right)=?$$$$2.\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\mathrm{sinx},\frac{\pi }{2}<\mathrm{x}⩽2\pi \\ \mathrm{cosx},0⩽\mathrm{x}⩽\frac{\pi }{2}\end{array}$$$$\mathrm{then}\mathrm{find}\mathrm{the}{\mathrm{f}}^{\prime }\left(2\pi \right)=?$$$$$$

Answered by mathlove last updated on 26/Jul/23

$$f^{{}^{\prime }}\left(x\right)=\{\begin{array}{l}cosx\\ -sinx\end{array}$$$$f^{{}^{\prime }}\left(\frac{\pi }{2}\right)=-sin\frac{\pi }{2}=-1$$$$f^{{}^{\prime }}\left(2\pi \right)=cos2\pi =1$$

Commented by MM42 last updated on 26/Jul/23

$$Thefunctionisnotcontinuousein$$$$x=\frac{\pi }{2}.thereforeitisnotdifferentiableinx=\frac{\pi }{2}$$

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